The rate of S_{N}2 reaction is: | vedclass

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(C) The $S_{N}2$ reaction requires a backside attack by the nucleophile on the carbon atom bonded to the leaving group. $(A)$ is $tert$-butyl bromide,

Vedclass · Accepted Answer

$(A) > (B) > (C)$

Vedclass · Answer

(C) The $S_{N}2$ reaction requires a backside attack by the nucleophile on the carbon atom bonded to the leaving group. $(A)$ is $tert$-butyl bromide,which is a tertiary halide. While $S_{N}2$ is generally slow for tertiary halides due to steric hindrance,it is still more accessible than the bridgehead carbons in bicyclic systems. $(B)$ is $1$-bromobicyclo$[2.2.2]$octane and $(C)$ is $1$-bromobicyclo$[2.2.1]$heptane. In these bicyclic compounds,the bridgehead carbon is extremely sterically hindered,and a backside attack is geometrically impossible because the rigid cage structure prevents the nucleophile from approaching the carbon from the opposite side of the $C-Br$ bond. Furthermore,the transition state for an $S_{N}2$ reaction requires a planar arrangement of the bonds around the central carbon,which cannot be achieved in these rigid bicyclic systems. Therefore,$(A)$ reacts the fastest,followed by $(B)$,and $(C)$ is the slowest due to the higher strain and rigidity of the $[2.2.1]$ system compared to the $[2.2.2]$ system. Thus,the order of reactivity is $(A) > (B) > (C)$.

The rate of $S_{N}2$ reaction is:

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